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# ABCD is a square & angle PQR=90

Asked by 1st March 2013, 12:44 AM
Answer : Given : ABCD is a square & angle PQR=90 .
P,Q,R are 3 points on AB,BC & CD respectively.AP=CR.
To Prove :
1.PB = QC
2.PQ=QR
3.Angle QPR=45

Let as assume that angle QPR =a and angle QRP = b

Now , in triangle PQR and triangle QCR
angle PQR = angle QCR = 90
angle PRQ = angle RQC = b .....(1) { opposite angles}
angle QPR = angle QRC = a....(2) {exterior angles}

therefore the triangles are congruent by AAA property.
=> QR= QC .....(3)
similarily triangle PQR is congruent to triangle PBQ  and as well congruent to triangle QCR
=> QC = QR = PB        {using 3}
also QR = PQ     {of triangle QCR and triangle PBQ } ......(4)

now eq 4 suggests that in triangle PQR PQ = QR
therefore triangle PQR is issoceles triangle
=> angle a = angle b .....(5)      {angles of the issoceles triangles                                                   are equal  }

Now in triangle PQR
90 + angle a+ angle b = 180
angle a+ angle b = 180 - 90 = 90
2 angle a = 90                 {using eq5}
=> angle a = 45 degree
i.e angle QPR = 45 degree
Hence proved
Answered by Expert 6th March 2013, 6:16 PM
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